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STRUCTURE OF Fe-56, Fe-55 AND Mn-55
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF STABLE Fe-56 WITH S =0 ' After a detailed analysis of the high symmetry of the structure of Cr-48 one concludes that the structures of the Fe-56 , Fe-55 and Mn-55 have the same core of the structure of Cr-48. (See my STRUCTURE OF Cr-52 AND Cr-48 ). Although the isolated Cr-48 has an unstable structure, the high symmetry of Cr-48 plays the role of the very stable core for the structure of Fe-56. In the diagram you see the compound core consisting of the structure of ca-40 with S=0 in which we added the additional deuterons like the p21n21 of S=-1, the p22n22 of S=+1, the p23n23 of S=-1 and the p24n24 of S=+1. That is, this core having the structure of Gr-48 has S=0. In order to get the stable structure of fe-56 with S=0 in the core we added the two additional deuterons like the p25n25 of S=-1 and the p26n26 of S=+1. Then the extra four neutrons which contribute to the stable structure of Fe-56 fill the following blank positions. At n27 you see the n27(+1/2) which makes the two np bonds like the (n27-p1) and the (n27-p21). At n28 there exists the n28(-1/2) which is not shown because it is in front of p14. Similarly at n29 the extra neutron n29(+1/2) is not shown because it is in front of p15. Then the fourth extra neutron like the n30(-1/2) fills the blank position between the p11 and p24. Such extra neutrons with opposite spins contribute not only to the high symmetry of the structure of Fe-56 with S= but also to the stability of Fe56 because they increase the pn bonds per proton able to overcome the pp repulsions of long range. ' STABLE STRUCTURE OF Fe-56 WITH S = 0 (The nucleons p19, n19, n20, and p20 are not shown here because they are behind the n6, p6, p8, and n8 respectively. Also the n17, p17, p18, and n18 are not shown here because they are in front of p5, n5, n7, and p7 respectively. Note that the extra neutrons like n28 and n29 are not shown because they fill the blank positions in front of p14 and p15 respectively) ' ' ' ' n25……….p12..........n12 ' p25………..n11..........p11 ……….n30 Sixth horizontal plane of 7 nucleons' ' p22..........n10..........p10..........n24' ' n22.........p9............n9..........p24 Fifth horizontal plane of 8 nucleons' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 Fourth horizontal plane of 8 nucleons' ' p13.............n6...........p6............n15' ' n13..........p5...........n5............p15 Third horizontal plane of 8 nucleons' ' n21……,,,,,p4............n4……….p23' ' p21………n3..........p3 ……… n23 Second horizontal plane of 8 nucleons' ' n2............p2……….n26' ' n27……….p1...........n1 ……….p26 First horizontal plane of 7 nucleons' STRUCTURE OF THE UNSTABLE Fe-55 WITH S = -3/2 Here in the absence of an extra neutron the pp repulsions of long range are able to overcome the pn bonds and lead to the decay of Fe-55. (It decays to the Mn-55). In the diagram you see that the deuteron p26n26 belongs to the nucleons not of the second horizontal plane as in the case of Fe-56 but to the nucleons of the sixth horizontal plane. Here the two extra nucleons like the n27(+1/2) and the n28(+1/2) fill the blank positions between the p1 and p21 as well as the p2 and p23. Whereas the n29(-1/2) is not shown here because it fills the blank position in front of the p14. In other words comparing this structure with the structure of Cr-48 of S=0 one concludes that the spin S= -3/2 of Fe-55 is due to the additional spins of the two deuterons like the p25n25 of S=-1 and the p26n26 of S =-1 and the summation of the spins of n27 (+1/2), n28(+1/2) and n29(-1/2). Thus we get S = -1 -1 +1/2 +1/2 -1/2 = -3/2. ' UNSTABLE STRUCTURE OF Fe-55 WITH S = -3/2' The deuterons of the alpha particles at the center of Mg-24 like the n17p17 and p18n18 are not shown here because they are in front of p5n5 and n7p7 respectively. Also the deuterons p19n19 and n20p20 are not shown here because they are behind the n6p6 and p8n8 respectively. Moreover the extra n29(-1/2) is not shown because it fills the blank positions in front of p14.. It brakes the high symmetry ad makes two weak horizontal bonds per neutron unable to overcome the pp and nn repulsions. ' ' ' n25………p12..........n12………p26' ' p25……….n11..........p11 ……n26 Sixth horizontal plane of 8 nucleons' ' p22.........n10..........p10..........n24' ' n22..........p9............n9..........p24 Fifth horizontal plane of 8 nucleons' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 Fourth horizontal plane of 8 nucleons' ' p13.............n6...........p6............n15' ' n13..........p5...........n5............p15 Third horizontal plane of 8 nucleons' ' n21……….p4............n4……….p23' ' p21………n3...........p3 ……… n23 Second horizontal plane of 8 nucleons' ' n2............p2……….n28' ' n27……….p1...........n1 First horizontal plane of 6 nucleons' STABLE STRUCTURE OF Mn-55 WITH S = -5/2 For the description of the following diagram of Mn-55 we used again the core of Cr-48. Now we get the stable structure of Mn-55 by adding the deuteron p25n25 with S= -1 . The additional deuteron is not shown here because its nucleons exist in front of the n3 and p3 respectively. Here the five extra neutrons like the n26(-1/2), the n27(-1/2), the n28(+1/2) the n29(-1/2) and the n30(-1/2 ) give a total spin S = -3/2. Thus adding the spin S=-1 of the additional deuteron we get the total spin of Mn-55 as S =-5/2. HOW THE UNSTABLE Fe-55 OF S =-3/2 DECAYS TO THE STABLE Mn-55 OF S = -5/2 ''' Under the instability of Fe-55 the deuteron p25n25 of S =-1 goes in the new structure of Mn-55 to fill the blank positions in front of n3 and p3 as p25n25 with the same S= -1. At the same time thep26 (-1/2) of Fe-55 decays to n27(-1/2) of Mn-55, while the n27(+1/2) of the Fe-55 goes to the second plane in front of p21 as n30(-1/2). Of course this change of spin gives S= -1 . Under this condition we get S = -3/2 -1 = -5/2. '''STABLE STRUCTURE OF Mn-55 WITH S =-5/2 (The nucleons p19, n19, n20, and p20 are not shown here because they are behind the n6, p6, p8, and n8 respectively. Also the n17, p17, p18, and n18 are not shown here because they are in front of p5, n5, n7, and p7 respectively. Similarly the additional deuteron like the p25n25 is not shown here because it is in front of p3 and n3. Moreover the extra neutrons like n29 and n30 are not shown because they fill the blank positions in front of p14 and p21 respectively) ' ' ' n27………p12..........n12' ' n11..........p11 ……n26 Sixth horizontal plane of 8 nucleons' ' p22...........n10..........p10..........n24' ' n22..........p9............n9..........p24 Fifth horizontal plane of 8 nucleons' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 Fourth horizontal plane of 8 nucleons' ' p13.............n6...........p6............n15' ' n13..........p5...........n5............p15 Third horizontal plane of 8 nucleons' ' n21……….p4............n4………..p23' ' p21………n3..........p3 ……… n23 Second horizontal plane of 8 nucleons' ' n2............p2……….n28' ' p1...........n1 First horizontal plane of 7 nucleons' Category:Fundamental physics concepts